I have a question about an equality of the expoents we are asked to show.

The book states: $||f||_r <= ||f||_p * \mu (X)^s$ where $s=1/r - 1/p$ when $\mu (X) < \infty $

The thing is when using holder inequation im getting $\mu (X)^{\dfrac{1}{r-p}}$ and its driving me crazy that i can't show ${\dfrac{1}{r-p}} =s$

The only relation i have is the one i've used for holder's inequation, and its $p/r + (r-p)/r = 1$

  • $\begingroup$ You didn't just write $\dfrac 1r - \dfrac 1s = \dfrac1{r-p}$, did you? $\endgroup$ – Ted Shifrin Jun 14 at 23:38
  • $\begingroup$ Nop, i have done some Dumb maths, Just don't know where $\endgroup$ – Daniel Moraes Jun 15 at 10:54

$\int |f|^{r} \leq (\int |f|^{p})^{r/p}(\mu(X))^{1-r/p}$ so $\|f||_r \leq\|f\|_p (\mu(X))^{s}$. I have applied Holder's inequality with indices $\frac p r$ and $\frac p {p-r}$.

  • $\begingroup$ I messed up my expoents thx! $\endgroup$ – Daniel Moraes Jun 15 at 10:55

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